leetcode42. Trapping Rain Water

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Given n non-negative integers representing an elevation map where the width of each bar is 1, compute how much water it is able to trap after raining.

For example, 
Given [0,1,0,2,1,0,1,3,2,1,2,1], return 6.

The above elevation map is represented by array [0,1,0,2,1,0,1,3,2,1,2,1]. In this case, 6 units of rain water (blue section) are being trapped. Thanks Marcosfor contributing this image!

 題目如上。假期最後一道hard難度的題，先放上之前實現的代碼，還可以簡化

 1 class Solution { 2 public: 3     int trap(vector<int>& height) { 4         int sum=0; 5         for(int i=0;i<height.size();++i){ 6             for (int j=i+1;j<height.size();++j) { 7                 if(height[j]>=height[i]){ 8                     sum+=height[i]*(j-i-1); 9                     //可最佳化，m->i10                     for(int m=i+1;m<j;++m){11                         sum-=height[m];12                     }13                     i=j-1;14                     break;15                 }16                 if(j==height.size()-1){17                     int max=height[i+1];18                     int max_i=i+1;19                     for(int n=max_i+1;n<height.size();++n){20                         if(height[n]>max){21                             max=height[n];22                             max_i=n;23                         }24                     }25                     sum+=height[max_i]*(max_i-i-1);  26                     for(int m=i+1;m<max_i;++m){27                         sum-=height[m];28                     }29                     i=max_i-1;30                 }31             }32 33         }34         return sum;35     }36 };

 

 

 

leetcode42. Trapping Rain Water